The data that support the findings of this study are available within the article.

Proof of Theorem  REF
With Condition \(C1\) , \([A_i, U^{\dag }B_jU] = 0\) . Thus, \([\sqrt{A_i}, U^{\dag }B_jU] = 0\) . The right hand side of (REF ) becomes \(Tr(U^{\dag }B_jUA_i\rho )\) . Given the completeness of POVM operators, \(\sum _iA_i = I\) , we have \(\sum _i Tr(U^{\dag }B_jUA_i\rho ) = Tr(U^{\dag }B_jU\rho )=Tr(B_j\rho (t_2))=p(b_j\text{ }at\text{ }t_2|\rho )\) .
Given Condition \(C2\) , \(\rho =\sum _k\lambda _k|\phi _k\rangle \langle \phi _k|\)  and \(A_i=|\phi _i\rangle \langle \phi _i|\) , we get \(\sqrt{A_i}\rho \sqrt{A_i}=\lambda _i|\phi _i\rangle \langle \phi _i|\) . Then the right hand side of (REF ) becomes \(Tr(U^{\dag }B_jU\lambda _i|\phi _i\rangle \langle \phi _i|)=\lambda _i\langle \phi _i|U^{\dag }B_jU|\phi _i\rangle \) . The right hand side of (REF ) becomes \(\sum _i\lambda _i\langle \phi _i|U^{\dag }B_jU|\phi _i\rangle \) . On the other hand, \(p(b_j\text{ }at\text{ }t_2|\rho )=Tr(U^{\dag }B_jU\rho )=\sum _k Tr(U^{\dag }B_jU\lambda _k|\phi _k\rangle \langle \phi _k|)=\sum _i\lambda _i\langle \phi _i|U^{\dag }B_jU|\phi _i\rangle \) , same as the right hand side of (REF ).
Given \(\rho =|\Psi \rangle \langle \Psi |\) , and \(A_i\)  is a projection operator, in Condition \(C3\) , we have \(A_i^2=A_i\) , and the right hand side of (REF ) becomes \(\sum _i\langle \Psi |A_iU^{\dag }B_jUA_i|\Psi \rangle \) . The left hand side \(p(b_j\text{ }at\text{ }t_2|\rho )=\langle \Psi |U^{\dag }B_jU|\Psi \rangle \) . Again, by the completeness of POVM operators, \(\sum _iA_i = I\) , and we get \(p(b_j\text{ }at\text{ }t_2|\rho )=\sum _i\langle \Psi |U^{\dag }B_jUA_iA_i|\Psi \rangle \) . To have both sides of (REF ) equal, we need
\(\sum _i\langle \Psi |A_iU^{\dag }B_jUA_i|\Psi \rangle - \sum _i\langle \Psi |U^{\dag }B_jUA_iA_i|\Psi \rangle = 0\) . This can be rearranged to \(\sum _i\langle \Psi |[A_i, U^{\dag }B_jU]A_i|\Psi \rangle = 0\) , and Condition \(C3\)  ensures this is the case.

Proof of (REF )
To avoid confusion, we need to restore the subscripts of operators as \(C_i\)  and \(D_j\) . Since \(C_i\)  and \(D_j\)  are projection operators, and \(\rho =|\Psi \rangle \langle \Psi |\) , the R.H.S. of (REF ) becomes \(\sum _{ij}\langle \Psi |C_iD_jABC_iD_j|\Psi \rangle \) . Given the completeness of \(\lbrace C_i\rbrace \)  and \(\lbrace D_j\rbrace \) , we have \(\sum _{ij}C_iD_j=I\) . Since \(C_i^2=C_i\)  and \(D_j^2=D_j\) , we further obtain \(\sum _{ij}C^2_iD^2_j=I\) . Then \(p(ab \text{ }at\text{ }t_2|\rho , xy)=Tr(\rho AB)=\sum _{ij}\langle \Psi |ABC^2_iD^2_j|\Psi \rangle \) . To make this equal to the R.H.S. of (REF ), one condition is to have \(\langle \Psi |C_iD_jABC_iD_j|\Psi \rangle =\langle \Psi |ABC^2_iD^2_j|\Psi \rangle \) . This is equivalent to \(\langle \Psi |(C_iD_jABC_iD_j-ABC^2_iD^2_j)|\Psi \rangle =0\) . But \((C_iD_jABC_iD_j-ABC^2_iD^2_j)=[C_iD_j, AB]C_iD_j\) . Thus we have \(\langle \Psi |[C_iD_j, AB]C_iD_j|\Psi \rangle =0\) . Omitting the subscripts of \(C_i\)  and \(D_j\)  again gives (REF ).

Non-commutation of Operators in  [1]} 
The key characteristic of the EWF experiment is that the super-observer Alice (or Bob) performs measurements on the laboratory \(L_1\)  (or \(L_2\) ) as a whole. Thus the measurement operator acts on both the observed system and the friend in the lab. [1]} carefully chooses the operators as following. When \(x=1\) , Alice's measurement is represented as \(A(x=1)=|c\rangle \langle c|_{F_1}\otimes I_{s_1}\)  where \(c\)  is the outcome Charlie obtains from his measurement on \(s_1\)  and \(F_1\)  refers to Charlie himself. For \(x\in \lbrace 2,3\rbrace \) , Alice's measurement operator is \(A(x)=U_{L_1}(I_{F_1}\otimes E^x_{s_1})U_{L_1}^{-1}\) , where \(U_{L_1}^{-1}\)  is a unitary evolution that reverses the entanglement between \(F_1\)  and \(s_1\) , and \(E^x_{s_1}\)  is a positive operator on \(s_1\)  associated with outcome \(a\)  for measurement setting \(x\) . The operator associated with Charlie's measurement on \(s_1\) , according to  [1]}, is described by an unitary operator \(C(x) = U_{L_1}\)  from Alice's perspective. \(U_{L_1}\)  acts on the same Hilbert space \({\cal {H}}_{F_1}\otimes {\cal {H}}_{s_1}\)  and entangles \(s_1\)  and \(F_1\) . For \(x=1\) , \([A(x), C(x)]= |c\rangle \langle c|_{F_1}\otimes I_{s_1}U_{L_1} - U_{L_1}|c\rangle \langle c|_{F_1}\otimes I_{s_1}\ne 0\) , and for \(x\in \lbrace 2,3\rbrace \) , \([A(x), C(x)]=U_{L_1}(I_{F_1}\otimes E^x_{s_1}) - U^2_{L_1}(I_{F_1}\otimes E^x_{s_1})U_{L_1}^{-1} \ne 0\) .
As already pointed out [4]}, defining \(C(x)\)  as \(U_{L_1}\)  implies pre-measurement only with no measurement result, and leads to contradictions. An alternative choice of operation is that Charles performs a projection operation after the pre-measurement. This refines the definition of \(C(x)\)  to include both \(U_{L_1}\)  and a projection operation on \(s_1\) , i.e., \(C(x)=U^{\dag }_{L_1}(I_{F_1}\otimes |c\rangle \langle c|)U_{L_1}\) . With this refined definition, one can verify that \([A(x), C(x)]\ne 0\)  still true. Choosing \(C(x)=U^{\dag }_{L_1}(I_{F_1}\otimes |c\rangle \langle c|)U_{L_1}\)  implies operator \(C(x)\)  is from Charlie's point of view. This may not be the original intention in [1]}. However, the key point here is that either choice of \(C(x)\) , \([A(x), C(x)]\ne 0\) . The same analysis goes to operators \(B(y)\)  and \(D(y)\) , and the conclusion is that \([B(y), D(y)]\ne 0\)  for \(y\in \lbrace 1,2,3\rbrace \) .

Proof That (REF ) Does Not Hold
First we consider a simpler case that only Alice performs the two type of measurements and Bob does nothing. The law of total probability in this case can take the form of \(p(a_2|\rho )=\sum _{a_1=\lbrace 0,1\rbrace }p(a_2|a_1,\rho )p(a_1|\rho )\) . We will show this is true due to the fact that the selected operators \(A_1\)  and \(A_2\) , and wave function (REF ) together meet condition \(C6\) . To see this, substitute \(A_1=|\phi _i\rangle \langle \phi _i|_{L_1}\)  into condition \(C6\) , \(C6\)  becomes
\(\langle \varphi _i|[A_1,A_2]|\phi _i\rangle =0, \text{ where } |\varphi _i\rangle = \langle \Psi |\phi _i\rangle |\Psi \rangle ,\) 
Now consider the case \(a_1=0\) , where \(|\phi _0\rangle =|00\rangle _{L_1}\) . From (REF ) one can calculate

0|= 12(-2|00L2+2|11L2)
|0= |0|= 12|00L1

Then from (REF ) and dropping the unimportant factor of \(1/2\)  for \(|\varphi _0\rangle \) , we have for the case of \(a_1=0\)
\(\begin{split}\langle \varphi _0|[A_1,A_2]|\phi _0\rangle &= \langle 00|(-1)^{a_2}(|00\rangle \langle 11| - |11\rangle \langle 00|)|00\rangle \\&=0.\end{split}\) 
For the case of \(a_2=1\) , we can verify that \(|\varphi _1\rangle =|11\rangle _{L_1}\)  and
\(\begin{split}\langle \varphi _1|[A_1,A_2]|\phi _1\rangle &= \langle 11|(-1)^{1+a_2}(|00\rangle \langle 11| - |11\rangle \langle 11|)|00\rangle \\&=0.\end{split}\) 
Therefore, condition \(C6\)  is met with the choices of wavefunction and Alice's measurement operation. Similarly, if only Bob performs the two types of measurements and Alice does not perform any measurement, and let \(B_1=|\phi ^{\prime }_i\rangle \langle \phi ^{\prime }_i|_{L_2}\) , we can verify that
\(\langle \varphi ^{\prime }_i|[B_1,B_2]|\phi ^{\prime }_i\rangle =0, \text{ where } |\varphi ^{\prime }_i\rangle = \langle \Psi |\phi ^{\prime }_i\rangle |\Psi \rangle .\) 
Thus, \(p(b_2|\rho )=\sum _{b_1=\lbrace 0,1\rbrace }p(b_2|b_1,\rho )p(b_1|\rho )\)  holds true per Corollary REF .
However, when we consider both Alice and Bob perform the measurements \(\lbrace A_1, A_2\rbrace \)  and \(\lbrace B_1, B_2\rbrace \) , respectively, the situation is different. By replacing operators \(A, B, C, D\)  in (REF ) with operators \(A_2, B_2, A_1, B_1\) , (REF ) reads
\(\langle \Psi |[A_1B_1, A_2B_2]A_1B_1|\Psi \rangle =0.\) 
(REF ) and (REF ) together are not sufficient to ensure (REF ) is valid. Consequently, the law of total probability such as \(p(a_2b_2|\rho )=\sum _{a_1,b_1=\lbrace 0,1\rbrace }p(a_2b_2|a_1b_1,\rho )p(a_2b_2|\rho )\) , is not valid. Let's confirm this by direct calculation for the case \(a_2=0\)  and \(b_2=0\) , where the corresponding projection operators are

A1 =|ii|L1, B1 =|ii|L2
A2=12(|00L1+ |11L1)(00|L1+ 11|L1)
B2=12(|00L2+ |11L2)(00|L2+ 11|L2).

where \(|\phi _0\rangle =|00\rangle \)  and \(|\phi _1\rangle =|11\rangle \) . From (REF ), one can calculate that

a1,b1={0,1}p(a2b2|a1b1,)p(a2b2|)
=a1,b1={0,1}|A1B1A2B2A1B1|

 = i,j={0,1}||ij|2 i|A2|ij|B2|j
= 14.

Meanwhile, given \([A_2\otimes I_{L_2}, I_{L_1}\otimes B_2]=0\) , the joint probability \(p(a_2b_2|\rho )\)  is well-defined as

p(a2=0,b2=0|)=Tr(A2B2)=|A2B2|
=116||(|00s1C+ |11s1C)(|00s2D+ |11s2D)|2
=1822.

Thus, \(p(a_2b_2|\rho )\ne \sum _{a_1,b_1=\lbrace 0,1\rbrace }p(a_2b_2|a_1b_1,\rho )p(a_2b_2|\rho )\)  for the case of \(a_2=0\)  and \(b_2=0\) . For other values of \(a_2, b_2\in \lbrace 0, 1\rbrace \) , similar results can be calculated. Consequently, the law of marginal probability \(p(a_2b_2|\rho )=\sum _{a_1,b_1=\lbrace 0,1\rbrace }p(a_1b_1,a_2b_2|\rho )\)  does not hold, if we define the joint probability \(p(a_1b_1,a_2b_2|\rho )=p(a_2b_2|a_1b_1,\rho )p(a_2b_2|\rho )\) .
If we add another condition that the quantum state \(|\Psi \rangle \)  is chosen as a product state of Hilbert space \(\mathcal {H}_{L_1}\)  and \(\mathcal {H}_{L_2}\) , then together with (REF ) and (REF ), (REF ) becomes true. To see this, let \(|\Psi ^{\prime }\rangle =|\xi \rangle _{L_1}\otimes |\zeta \rangle _{L_2}\) , () becomes

 i,j={0,1}||i|j|2 i|A2|ij|B2|j

=i||i|2i|A2|ij||j|2j|B2|j.

But (REF ) implies \(\sum _{i}|\langle \xi |\phi _i\rangle |^2\langle \phi _i|A_2|\phi _i\rangle =\langle \xi |A_2|\xi \rangle \) , and (REF ) implies \(\sum _j|\langle \zeta |\phi _j\rangle |^2\langle \phi _j|B_2|\phi _j\rangle =\langle \zeta |B_2|\zeta \rangle \) . Thus, () becomes \(\langle \xi |A_2|\xi \rangle \langle \zeta |B_2|\zeta \rangle = \langle \Psi ^{\prime }|A_2B_2|\Psi ^{\prime }\rangle =p(a_2b_2|\rho )\) . This confirms (REF ) is valid.
However, \(|\Psi \rangle \)  in (REF ) is an entangled state between Hilbert space \(\mathcal {H}_{L_1}\)  and \(\mathcal {H}_{L_2}\) , so that (REF ) does not hold.
